\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|-3,2+\dfrac{2}{5}\right|\)

=>\(\left|x-\dfrac{1}{3}\right|+0,8=\left|-3,2+0,4\right|=2,8\)

=>\(\left|x-\dfrac{1}{3}\right|=2,8-0,8=2\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

đề sai

\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\left(-3,2\right)+\dfrac{2}{5}\right|\)

\(\Leftrightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{14}{5}\)

\(\Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{14}{5}-\dfrac{4}{5}\)

\(\Rightarrow\left|x-\dfrac{1}{3}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-5}{3}\end{matrix}\right.\)

Vậy..............

\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \left | \left ( - 3,2 \right ) + \frac{2}{5}\right |\)

\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \left | - \frac{14}{5} \right |\)

\(\left | x - \frac{1}{3} \right | + \frac{4}{5} = \frac{14}{5} \)

\(\left | x - \frac{1}{3} \right | = 2\)

* \(x - \frac{1}{3}= 2\)

x = 2 + \( \frac{1}{3}\)

\(x = \frac{7}{3}\)

* \(x - \frac{1}{3}= - 2\)

\(x = - 2 + \frac{1}{3}\)

\(x = - \frac{5}{3}\)

Vậy x = \(x = \frac{7}{3}; x = - \frac{5}{3}\)

\(\left\{\dfrac{-5< 0< -0,4}{x\in Z}\right\}\Rightarrow x\in\left\{-4;-3;-2;-1\right\}\)

chiu thôi leo nheo qua

chịu không viết được hỗn số

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

a) \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\frac{-16}{5}+\frac{2}{5}\right|\)

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\frac{-14}{5}\right|\)

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)

\(\left|x-\frac{1}{3}\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-5}{3}\end{cases}}\)

làm tiếp câu a) nhé

b) \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x-7=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8